Question

The solution set of the inequality $\left(\tan ^{-1} x\right)^{2} \leq\left(\tan ^{-1} x\right)+6$ is

• A. $\left[- tan 2 , tan ⁡ 3\right]$
• B. $\left[tan 2 , - tan ⁡ 3\right]$
• C. $\left[tan 2 , tan ⁡ 3\right]$
• D. $\left(- \in fty , \in fty\right)$

Answer 1

Answer: (D)

Let $\text{tan}^{- 1}x=u$
So, inequality becomes $u^{2}-u-6\leq 0$
$\Rightarrow -2\leq u\leq 3$
So, $-2\leq \text{tan}^{- 1}x\leq 3$
As $-\frac{\pi }{2} < \text{tan}^{- 1}x < \frac{\pi }{2}$
$\Rightarrow -2\leq \text{tan}^{- 1}x\leq 3, \, \forall x\in R$
Therefore, the solution set of the inequality $\left(\tan ^{-1} x\right)^{2} \leq\left(\tan ^{-1} x\right)+6$ is $(-\infty, \infty)$